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  <meta name="description" content="数学归纳法的原理： 为证明对所有的正整数n，p(n)为真，其中p(n)是一个命题函数 需要完成两个步骤     基础步骤： 证明命题p(1)成立     归纳步骤： 证明对每个正整数k来说， p(k) --&gt; p(k+1)      强归纳法： 要证明对所有的正整数n而言，都有p(n)为真，其中p(n)为命题函数，我们 要完成如下两个步骤：     基础步骤： 证明p(1)为真     归纳">
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          数学归纳法和强归纳法
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        <pre><code>数学归纳法的原理： 为证明对所有的正整数n，p(n)为真，其中p(n)是一个命题函数
需要完成两个步骤
    基础步骤： 证明命题p(1)成立
    归纳步骤： 证明对每个正整数k来说， p(k) --&gt; p(k+1)

    强归纳法： 要证明对所有的正整数n而言，都有p(n)为真，其中p(n)为命题函数，我们
要完成如下两个步骤：
    基础步骤： 证明p(1)为真
    归纳步骤： 要证明对所有正整数k来说，蕴含式 p(1)&amp;p(2)...p(k)  --&gt; p(k+1)也为真
</code></pre><a id="more"></a>    
<pre><code>example：
a) Determine which amounts of postage can be formed
using just 4-cent and 11-cent stamps.
b) Prove your answer to (a) using the principle of math-
ematical induction. Be sure to state explicitly your
inductive hypothesis in the inductive step.
c) Prove your answer to (a) using strong induction. How
does the inductive hypothesis in this proof differ from
that in the inductive hypothesis for a proof using math-
ematical induction?

a)、4,8,11,12,15,16,19,20,22,23,24,26,27,28，以及大于等于30的所有值
b)、令P(n)是命题：我们可以只用4分邮票和11分邮票组成n分邮资。证明：对于所有 n &gt;= 30, P(n)为真。
基础步骤：可以组成k分邮资，如何组成k+1分邮资
归纳步骤：如果k分邮资中有一张11分邮票，可以用3张4分邮资代替组成k+1；否则k分邮资只由4分邮票构成。因为k&gt;=30,可以用三张11分邮票代替8张4分邮票，组成k+1分邮资。
c)P(n)与b中相同。
基础步骤：P(30) = 11+11+4, p(31) = 11+4+4+4+4+4, p(32)=4+4+4+4+4+4+4+4, p(33) = 11+11+11;
归纳步骤：假设归纳假设p(j)对于所有j(30&lt;=j&lt;=k)都为真，这里k是大于等于33的任意整数。我们要证明p(k+1)为真。因为k-3&gt;=30,所以p(k-3)为真，即可以组成k-3分邮资。在这个信封上在贴一张4分邮票，就组成了k+1分邮资。在这个证明中，我们的归纳假设是p(j)对所有介于30和k之间的值都为真，而不只是p(30)为真。
</code></pre>
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